Find the exact value of tan(θ) given that sin(θ) = 3/5 and θ is in the second quadrant
Given that $ \sin(\theta) = \frac{3}{5} $ and $ \theta $ is in the second quadrant:
Since $ \sin(\theta) $ is positive in the second quadrant, $ \cos(\theta) $ must be negative:
Use the Pythagorean identity:
$$ \sin^2(\theta) + \cos^2(\theta) = 1 $$
Substitute $ \sin(\theta) = \frac{3}{5} $:
$$ \left(\frac{3}{5}\right)^2 + \cos^2(\theta) = 1 $$
$$ \frac{9}{25} + \cos^2(\theta) = 1 $$
$$ \cos^2(\theta) = 1 – \frac{9}{25} = \frac{16}{25} $$
Since $ \theta $ is in the second quadrant, $ \cos(\theta) $ is negative:
$$ \cos(\theta) = -\frac{4}{5} $$
Now find $ \tan(\theta) $:
$$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4} $$
Thus, $ \tan(\theta) = -\frac{3}{4} $.